grep with variable in a variable

shell grep quoting

12,577

Change

ligne=`cat /var/log/svlog | grep  "\$day"`

ligne=$(grep "$day" /var/log/svlog)

To feed the contents of $ligne to awk, use

echo "$ligne" | awk ...

12,577

LinuxUser

Updated on September 18, 2022

Comments

LinuxUser over 1 year
I want to filter the svlog file by date and split the result by the space, so when i enter the date, it didn't work for me, please refer to the script that I wrote below, the problem was in this command:
```
ligne=`cat /var/log/svlog | grep  "\$day"`
```
, it doesn't consider "\$day". i also tried this "^\$day", this "$day" and this "${day}" but the same result.
```
#!/bin/bash                 

echo SCRIPT-LOG             
echo enter date         
read day                   

ligne=`cat /var/log/svlog | grep  "\$day"`

$ligne >> log1.txt          

awk '{split($ligne,numbers," ")} END {for(n in numbers){ print numbers[n] }}'>lo


"Monit_Sub.sh" 11 lines, 211 characters

$ sudo ./Monit_Sub.sh 
SCRIPT-LOG
enter date
Apr 26
./Monit_Sub.sh: line 8: Apr: command not found

$ 
```
- LinuxUser about 8 years
  
  just to correct ligne=` cat /var/log/svlog | grep "\$day" ` with this quotes (`)
- EightBitTony about 8 years
  
  This line "$ligne >> log1.txt" is missing an echo so your script is trying to execute a command called Apr. That's the specific error you've got, I've not considered anything else in your script.
- LinuxUser about 8 years
  
  The problem is not on that line "$ligne >> log1.txt", i have omitted it . but the "ligne" variable is empty there is no result.
- terdon about 8 years
  
  Please edit your question and explain what you are actually trying to do. You are showing your attempt but we have no way of knowing what your actual objective is. For example, is $ligne >> log1.txt supposed to execute the command stored in $ligne and append its output into log1.txt (that's what it does) or was it supposed to append the contents of the variable $ligne to the file? Your awk command has no input, what is that supposed to do? Please edit and clarify.
- roaima about 8 years
  
  Please provide a sample data file and the desired output using that file. You should edit your question to include this information.
- Wildcard about 8 years
  
  Start using $(this syntax) for command substitution instead of backticks; the quoting is MUCH easier.
- Ciro Santilli Путлер Капут 六四事 almost 7 years
  
  unix.stackexchange.com/questions/163810/grep-on-a-variable
terdon about 8 years

Why would you echo that? Just grep "$day" /var/log/svlog >> log1.txt is enough.
Guido about 8 years

@terdon I put in an echo because it's not entirely clear to me what OP intended to do with $ligne; see @EightBitTony 's comment above
LinuxUser about 8 years

i dont want echo it. i want to use the result of awk '{split($ligne,numbers," ")} END {for(n in numbers){ print numbers[n] }}'>lo
Guido about 8 years

@LinuxUser I see. Updated my answer.
LinuxUser about 8 years

I dont want to echo the variable ligne, i have ommitted this line "$ligne >> log1.txt" . I want to split the result of the variable "ligne", but there is no result.
Guido about 8 years

To process the contents of $ligne with awk, you will need to echo it into a pipe to awk, see answer.
LinuxUser about 8 years

It works for me now, is the missing echo on the last line echo $ligne | awk '{split($0,numbers," ")} END {for(n in numbers){ print numbers[n] }}'